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The Number of symmetric,Positive Definite, $8\times 8$ matrices having trace$=8$ and determinant$=1$ is

  1. $0$

  2. $1$.

  3. $>1$ but finite.

  4. $\infty$

I am not able to do this one.

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Hint: diagonal matrices. –  Gerry Myerson Jul 8 '12 at 23:03
    
Yes, so Only $1$ –  El Angel Exterminador Jul 8 '12 at 23:17

1 Answer 1

up vote 3 down vote accepted

If $A$ is pos. def. then its eigenvalues $\lambda_i$ are real and positive. Besides, we know (don't we?) that, for any matrix, $\sum \lambda_i = tr(A)$ and $\prod \lambda_i= |A|$. In our case, that means that we are restricted to $\sum \lambda_i =8$ and $\prod \lambda_i =1$... (can you go on from here? hint: arithmetic-geometric means and their properties)

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Here AM=GM for $a_1,\dots,a_8$, does that mean $a_1=\dots=a_8$? –  El Angel Exterminador Jun 3 '13 at 6:30
1  
Yes, that the AM equals the GM implies that all terms are equal, which in this case implies that $\lambda_i=1$, which in turns implies that the matrix is the identity. –  leonbloy Jun 3 '13 at 13:51

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